This seems like hard problem. Anyone has any suggestion how to tackle it?
Show that in a finite incidence geometry, the number of lines is greater than or equal to the number of points.

Between any two distinct points there exists exactly one line. This is one of the axioms of incidence geometry. Now if there are points (and say ) then the number of pairs that we can form is equal to while the number of lines is .

Between any two distinct points there exists exactly one line. This is one of the axioms of incidence geometry. Now if there are points (and say ) then the number of pairs that we can form is equal to while the number of lines is .

Did you mean that the number pairs we can form is equal to which is the number lines, while the number of points is only ? I did learn the axiom, and thanks a lot for your help TPH. It seems much easier.

@TPH, actually, I tried to draw some examples for some n points, but it appears that the number lines can be less that . I guess it could be that is the maximum number of lines that we can get from n points. If this is the case, then I'm stuck again

Incidence geometry: #lines >= #points

In an incidence geometry, two points are connected by exactly one line, there are (at least) three noncollinear points, and every line contains at least two points. In this context the de Bruijn-Erdos theorem (1948) says that there are at least as many lines as points. There's a proof in the text Combinatorics of Finite Geometries by Lynn Margaret Batten. One can also find proofs on the net by searching for "de Bruijn-Erdos theorem".